How many answers do I have?

[Fullmetal_Hye]

Solve for x. -2pi<x<2pi

1) sin(x)cos(x)=1/4
2sin(x)cos(x)=1/2
sin2(x)=1/2
2x=pi/6, 5pi/6, ??

2) cos4x=rad(3)/2
4x=pi/6, ??

Mm I don't very well know which quadrants I should consider[/img]

 

[tkhunny]

If you HAVE 2x and you WANT \(\displaystyle \L\,-2\pi \le x \lt 2\pi\), you had better examin solutions here \(\displaystyle \L\,-4\pi \le 2x \lt 4\pi\)

 

[soroban]

Hello, Fullmetal_Hye!

Solve for $\displaystyle x,\;\;-2\pi \,\leq x\,\leq\,2\pi$

$\displaystyle 1)\;\sin x\cos x \:=\:\frac{1}{4}$
$\displaystyle 2\sin x\cos x \:=\:\frac{1}{2}$
$\displaystyle \sin2x \:=\:\frac{1}{2}$

$\displaystyle 2x\:=\:\frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{13\pi}{6},\;\frac{17\pi}{6}$

Therefore: $\displaystyle x \:=\:\frac{\pi}{12},\;\frac{5\pi}{12},\;\frac{13\pi}{12},\;\frac{17\pi}{12}$

$\displaystyle 2)\;\cos4x \:=\:\frac{\sqrt{3}}{2}$

$\displaystyle 4x\:=\:\frac{\pi}{6},\;\frac{11\pi}{6},\;\frac{13\pi}{6},\;\frac{23\pi}{6},\;\frac{25\pi}{6},\;\frac{35\pi}{6},\;\frac{37\pi}{6},\;\frac{47\pi}{6}$

Then: $\displaystyle x\:=\:\frac{\pi}{24},\;\frac{11\pi}{24},\;\frac{13\pi}{24},\;\frac{23\pi}{24},\;\frac{25\pi}{24},\;\frac{35\pi}{24},\;\frac{37\pi}{24},\;\frac{47\pi}{24}$

I don't very well know which quadrants I should consider.

I understand your problem: you don't know "how far to go".


If we had, for example, $\displaystyle \sin x\,=\,\frac{1}{2}$, we know that: \(\displaystyle x\:=\:\frac{\pi}{6},\;\frac{\5\pi}{6}\)

$\displaystyle \;\;$That is, we know there are two answers on the interval $\displaystyle [0,2\pi]$


If we had: $\displaystyle \sin$2$\displaystyle x\,=\,\frac{1}{2}$, we would "go around twice".

$\displaystyle \;\;$We'd have: $\displaystyle \,2x\:=\:\frac{\pi}{6},\;\frac{5\pi}{6}$ and $\displaystyle \frac{13\pi}{6},\;\frac{17\pi}{6}$
(We get the last two by adding $\displaystyle 2\pi$ to the first two.)

Solve for $\displaystyle x$ (divide by 2): $\displaystyle \,x\;=\;\frac{\pi}{12},\;\frac{5\pi}{12},\;\frac{13\pi}{12},\;\frac{17\pi}{12}$
(Note that the last one is still less than $\displaystyle 2\pi$.)


If we have: $\displaystyle \sin$3$\displaystyle x\,=\,\frac{1}{2}$, we would "go around three times:.

$\displaystyle \;\;$We'd have: $\displaystyle \,\frac{\pi}{6},\;\frac{5\pi}{6}$ and $\displaystyle \frac{13\pi}{6},\;\frac{17\pi}{6}$ and $\displaystyle \frac{25\pi}{6},\;\frac{29\pi}{6}$

Solve for $\displaystyle x$ (divide by 3): $\displaystyle \,x\:=\:\frac{\pi}{18},\;\frac{5\pi}{18},\;\frac{13\pi}{18},\;\frac{17\pi}{18},\;\frac{25\pi}{18},\;\frac{29\pi}{18}$


See how it works?